Problem: $ E = \left[\begin{array}{rr}2 & 1 \\ 2 & -2\end{array}\right]$ $ F = \left[\begin{array}{rr}5 & 1 \\ 2 & 1\end{array}\right]$ What is $ E F$ ?
Answer: Because $ E$ has dimensions $(2\times2)$ and $ F$ has dimensions $(2\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ E F = \left[\begin{array}{rr}{2} & {1} \\ {2} & {-2}\end{array}\right] \left[\begin{array}{rr}{5} & \color{#DF0030}{1} \\ {2} & \color{#DF0030}{1}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ E$ , with the corresponding elements in column $j$ of the second matrix, $ F$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ E$ with the first element in ${\text{column }1}$ of $ F$ , then multiply the second element in ${\text{row }1}$ of $ E$ with the second element in ${\text{column }1}$ of $ F$ , and so on. Add the products together. $ \left[\begin{array}{rr}{2}\cdot{5}+{1}\cdot{2} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ E$ with the corresponding elements in ${\text{column }1}$ of $ F$ and add the products together. $ \left[\begin{array}{rr}{2}\cdot{5}+{1}\cdot{2} & ? \\ {2}\cdot{5}+{-2}\cdot{2} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ E$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ F$ and add the products together. $ \left[\begin{array}{rr}{2}\cdot{5}+{1}\cdot{2} & {2}\cdot\color{#DF0030}{1}+{1}\cdot\color{#DF0030}{1} \\ {2}\cdot{5}+{-2}\cdot{2} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{2}\cdot{5}+{1}\cdot{2} & {2}\cdot\color{#DF0030}{1}+{1}\cdot\color{#DF0030}{1} \\ {2}\cdot{5}+{-2}\cdot{2} & {2}\cdot\color{#DF0030}{1}+{-2}\cdot\color{#DF0030}{1}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}12 & 3 \\ 6 & 0\end{array}\right] $